If the equation of an elipse is 3x²+2y²+6x-8y+5=0, then which of the following are true?

Correct Answer is : foci are (-1,1) and (-1,3)

Correct Answer is : foci are (-1,1) and (-1,3)

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VITEEE 2010 Paper

Question : 104 of 120

Marks: +1, -0

3x^{2}+2y^{2}+6x-8y+5=0,

then which of the following are true?

Solution:

The equation of ellipse is

3x^{2}+2y^{2}+6x-8y+5=0

⇒

3(x^{2}+2x)+2(y^{2}-4y)+5=0

⇒

3(x^{2}+2x+1)+2(y^{2}-4y+4)+5-3-8=0

\Rightarrow 3(x+1)^{2}+2(y-2)^{2}=6

\Rightarrow\frac{(x+1)^{2}}{2}+\frac{(y-2)^{2}}{3}=1

Comparing with

\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1,

we get

h = -1, k = 2, a^{2}=2, b^{2}=3

Here, centre (h , k) = (-1, 2)

And using

a^{2}=b^{2}(1-e^{2})

2=3(1-e^{2})\Rightarrow e=\frac{1}{\sqrt{3}}

And foci are (h , k + be) and (h , k - be)

=(-1,2+1)\text{ and }(-1,2-1)

=(-1,3)\text{ and }(-1,1)

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